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[crumbs22] Week 04 Solution #1372

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36 changes: 36 additions & 0 deletions coin-change/crumbs22.cpp
Original file line number Diff line number Diff line change
@@ -0,0 +1,36 @@
#include <vector>
#include <algorithm>
using namespace std;

/*
- dp[i]: 금액 i를 만드는 최소 코인수
- 초기값
- dp[0] = 0
- 그 외에는 amount + 1 (도달할 수 없는 값)으로 초기화
- 현재 금액 i를 만들기 위해서
이전 금액인 i - c를 만들고, 거기에 코인 c를 더 썼을 때 최소값을 갱신함
예시) coins = [1, 2, 5], i = 3일 때
c = 1 -> 3 >= 1 이므로 dp[3] = min(dp[3], dp[2] + 1) 갱신
c = 2 -> 3 >= 2 이므로 dp[3] = min(dp[3], dp[1] + 1) 갱신
c = 5 -> 3 < 5 이므로 건너뜀
=> i - c가 음수면 배열 범위를 벗어나므로 i >= c일때만 연산산
*/
class Solution {
public:
int coinChange(vector<int>& coins, int amount) {
// dp 테이블 초기화
vector<int> dp(amount + 1, amount + 1);
dp[0] = 0;

// bottom up 연산
for (int i = 1; i <= amount; ++i) {
for (int c : coins) {
if (i >= c) {
dp[i] = min(dp[i], dp[i - c] + 1);
}
}
}

return (dp[amount] > amount) ? -1 : dp[amount];
}
};
32 changes: 32 additions & 0 deletions find-minimum-in-rotated-sorted-array/crumbs22.cpp
Original file line number Diff line number Diff line change
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#include <iostream>
#include <vector>

using namespace std;

/*
O(logn)으로 풀어라
- 이진 탐색 느낌
- mid값과 오른쪽 끝을 비교,
- mid > 오른쪽 끝 : 최소는 오른쪽에
- mid <= 오른쪽 끝 : 최소는 mid 포함 왼쪽에
*/

class Solution {
public:
int findMin(vector<int>& nums) {
int left = 0;
int right = nums.size() - 1;

while (left < right) {
int mid = left + (right - left) / 2;

if (nums[mid] > nums[right]) {
left = mid + 1;
}
else {
right = mid;
}
}
return (nums[left]);
}
};
19 changes: 19 additions & 0 deletions maximum-depth-of-binary-tree/crumbs22.cpp
Original file line number Diff line number Diff line change
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#include <algorithm>

struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
TreeNode() : val(0), left(nullptr), right(nullptr) {}
TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
};

class Solution {
public:
int maxDepth(TreeNode* root) {
if (!root)
return (0);
return (std::max(maxDepth(root->left), maxDepth(root->right)) + 1);
}
};
44 changes: 44 additions & 0 deletions merge-two-sorted-lists/crumbs22.cpp
Original file line number Diff line number Diff line change
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#include <iostream>

using namespace std;

struct ListNode {
int val;
ListNode *next;
ListNode() : val(0), next(nullptr) {}
ListNode(int x) : val(x), next(nullptr) {}
ListNode(int x, ListNode *next) : val(x), next(next) {}
};

class Solution {
public:
ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
if (!list1)
return (list2);
if (!list2)
return (list1);

ListNode ans_head = ListNode();
ListNode* tmp = &ans_head;

while (list1 && list2) {
if (list1->val <= list2->val) {
tmp->next = list1;
list1 = list1->next;
tmp = tmp->next;
}
else {
tmp->next = list2;
list2 = list2->next;
tmp = tmp->next;
}
}
if (list1) {
tmp->next = list1;
}
else if (list2) {
tmp->next = list2;
}
return (ans_head.next);
}
};
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c++ 코드는 오랜만에 보는데 java와 다르게 포인터를 써서 주소값을 tmp에 넣어서 하는 것이 차이점으로 보이네요!

41 changes: 41 additions & 0 deletions word-search/crumbs22.cpp
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#include <vector>
#include <string>

using namespace std;

class Solution {
public:
bool exist(vector<vector<char>>& board, string word) {
// 단어 시작점을 탐색, 시작점 찾으면 dfs 시작하고 마지막 단어까지 도달 가능할 때 true 반환
for (int i = 0; i < board.size(); i++) {
for (int j = 0; j < board[0].size(); j++) {
if (board[i][j] == word[0] && dfs(board, i, j, 0, word)) {
return (true);
}
}
}
return (false);
}
bool dfs(vector<vector<char>>& board, int i, int j, int idx, string& word) {
if (idx == word.size())
return (true);
if (i < 0 || i >= board.size() || \
j < 0 || j >= board[0].size() || \
board[i][j] != word[idx]) {
return (false);
}

char tmp = board[i][j];

// 방문 표시
board[i][j] = '1';

// 4방향으로 탐색
bool found = dfs(board, i + 1, j, idx + 1, word) || dfs(board, i - 1, j, idx + 1, word) || \
dfs(board, i, j + 1, idx + 1, word) || dfs(board, i, j - 1, idx + 1, word);

// 다른 경로 탐색을 위해서 방문 표시를 원래대로 복원
board[i][j] = tmp;
return (found);
}
};